Chapter 1.2, Problem 4GCE

To determine

To find: $Y_3$ and use table to observe the result for definition of $Y_3$ .

Answer to Problem 4GCE

  $\begin{array}{c}{Y}_3=Y_1+Y_2=x^2-x+4\\ Y_3=Y_1-Y_2=x^2+x-6\\ Y_3=Y_1\cdot Y_2=-x^3+5x^2+x-5\\ Y_3=\frac{Y_1}{Y_2}=\frac{x^2-1}{5-x}\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}{Y}_1=f\left(x\right)=x^2-1\\ Y_2=f\left(x\right)=5-x\\ Y_3=Y_1+Y_2\\ Y_3=Y_1-Y_2\\ Y_3=Y_1\cdot Y_2\\ Y_3=\frac{Y_1}{Y_2}\end{array}$

The table of operation with function is as follows:

Operation	Mathematical expression for operation
Sum	$\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$
Difference	$\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$
Product	$\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$
Quotient	$\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$

Calculation:

Let’s substitute the value of $Y_1\text{ and }{Y}_2$ in the sum equation for $Y_3$ and simplify as follows:

  $\begin{array}{c}{Y}_3=Y_1+Y_2\\ =\left(x^2-1\right)+\left(5-x\right)\\ =x^2-x+4\end{array}$

Thus, the value of $Y_1+Y_2$ is $x^2-x+4$ .

Let’s substitute the value of $Y_1\text{ and }{Y}_2$ in the difference equation for $Y_3$ and simplify as follows:

  $\begin{array}{c}{Y}_3=Y_1-Y_2\\ =\left(x^2-1\right)-\left(5-x\right)\\ =x^2-1-5+x\\ =x^2+x-6\end{array}$

Thus, the value of $Y_1-Y_2$ is $x^2+x-6$ .

Let’s substitute the value of $Y_1\text{ and }{Y}_2$ in the product equation for $Y_3$ and simplify as follows:

  $\begin{array}{c}{Y}_3=Y_1\cdot Y_2\\ =\left(x^2-1\right)\cdot \left(5-x\right)\\ =5x^2-5-x^3+x\\ =-x^3+5x^2+x-5\end{array}$

Thus, the value of $Y_1\cdot Y_2$ is $-x^3+5x^2+x-5$ .

Let’s substitute the value of $Y_1\text{ and }{Y}_2$ in the quotient equation for $Y_3$ and simplify as follows:

  $\begin{array}{c}{Y}_3=\frac{Y_1}{Y_2}\\ =\frac{x^2-1}{5-x}\end{array}$

Thus, the value of $\frac{Y_1}{Y_2}$ is $\frac{x^2-1}{5-x}$ .